Lemma: If x+1x=2cosα, then xn+1xn=2cosnα
If you want to see the proof of this, go here. I have done the proof in a previous answer before.
Now, if we consider α=2π3 and n=99, using the formula directly gives
x99+1x99=2cos(2π3⋅99)=2
.An easy-to-understand solution:
After writing the terms, let S be the original sum and consider S/3.
S | = | 1/3 | + | 2/9 | + | 3/27 | + | ... |
S/3 | = | 1/9 | + | 2/27 | + | ... |
Subtracting gives:
S | = | 1/3 | + | 2/9 | + | 3/27 | + | ... |
S/3 | = | 1/9 | + | 2/27 | + | ... | ||
2S/3 | = | 1/3 | + | 1/9 | + | 1/27 | + | ... |
Now 2S/3 is the sum of geometric series 1/3 + 1/9 + 1/27 + ...
2S3=131−13=12S=34
Therefore the required sum is 3/4.